A student argues that ∫₁^∞ (1/x) dx must converge because 1/x → 0 as x → ∞, so the 'area added' eventually becomes negligible. What is wrong with this reasoning?
AThe student is correct — 1/x → 0 guarantees the integral converges
BThe integrand going to zero is necessary but not sufficient — 1/x decays too slowly for the total area to remain finite
CThe antiderivative of 1/x does not exist, so the integral cannot be evaluated
DThe integral should start at 0, not 1, to be a proper improper integral
This is the central misconception. The integrand 1/x does go to zero, but it goes to zero too slowly — its antiderivative is ln(x), which grows without bound. The p-test makes this precise: ∫₁^∞ 1/xᵖ dx converges if and only if p > 1. For p = 1, the 'area added' accumulates faster than it shrinks, so the total is infinite. Convergence requires the decay to be fast enough, not merely present.
Question 2 Multiple Choice
Which of the following improper integrals converges?
A∫₁^∞ (1/x) dx
B∫₁^∞ (1/√x) dx
C∫₁^∞ (1/x²) dx
D∫₁^∞ (1/(x ln x)) dx
By the p-test, ∫₁^∞ 1/xᵖ dx converges if and only if p > 1. Option A has p = 1 (diverges). Option B has p = 1/2 (diverges). Option C has p = 2 > 1 (converges to 1). Option D diverges — its antiderivative is ln(ln x), which grows without bound. Only C meets the p > 1 threshold.
Question 3 True / False
The integral ∫₋₁¹ (1/x) dx equals 0, because 1/x is an odd function and the interval [−1, 1] is symmetric about 0.
TTrue
FFalse
Answer: False
This is a dangerous misconception. The function 1/x has an infinite discontinuity at x = 0, making this an improper integral that must be evaluated as a limit. When you do so carefully (using the Cauchy principal value framework or splitting into two one-sided limits), both ∫₋₁⁰ (1/x) dx and ∫₀¹ (1/x) dx diverge. The symmetry argument cannot be applied because the integral does not exist — you cannot use antisymmetry to cancel two divergent quantities. Naively applying the FTC gives the numerically wrong answer of 0.
Question 4 True / False
Every improper integral must be evaluated as a limit — you cannot simply plug in ∞ or a discontinuity point directly.
TTrue
FFalse
Answer: True
This is the foundational procedure. ∫ₐ^∞ f(x) dx is defined as lim_{b→∞} ∫ₐᵇ f(x) dx; if this limit exists and is finite, the integral converges. For a discontinuity at an endpoint, e.g., ∫₀¹ 1/√x dx, you write lim_{a→0⁺} ∫ₐ¹ 1/√x dx. Plugging in ∞ directly has no mathematical meaning — ∞ is not a real number, so arithmetic with it is undefined. The limit formulation is what makes the definition rigorous.
Question 5 Short Answer
Explain why the p-integral ∫₁^∞ 1/xᵖ dx converges when p > 1 but diverges when p ≤ 1, even though the integrand goes to zero in all cases.
Think about your answer, then reveal below.
Model answer: When p > 1, the antiderivative x^(1−p)/(1−p) has 1−p < 0, so it goes to 0 as x → ∞, giving a finite limit. When p = 1, the antiderivative is ln x, which grows without bound. When p < 1, the antiderivative also grows without bound. The integrands all go to zero, but the rate of decay determines whether the accumulated area stays finite. Fast enough decay (p > 1) means the infinite tail has finite total area; too-slow decay means the accumulation wins.
The key is that going to zero is not enough — the function must go to zero fast enough. The p-test quantifies 'fast enough' precisely: p must exceed 1. This also explains why harmonic series and the integral of 1/x are the boundary cases that just barely diverge. Connecting this to series (the integral test) reveals the same threshold: the harmonic series diverges precisely because ∫₁^∞ 1/x dx diverges.